math problem?
If the average (arithmetic mean) of the first 16 positive integers is subtracted from the average (arithmetic mean) of the next 16 positive integers, what is the result?
128
64
32
16
0
In a geometric series, each term is a constant multiple of the preceding one. If the first three terms in a geometric series are -2, x, and -32, which of the following would be the 6th term in the series?
-4,096
-1,024
512
1,024
2,048
Comments
First question the answer is 16
Second answer (not sure) but I believe it is a positive digit, 2048.
The first poster is right. The answers are 16 and 2048
To see this, the aritmetic mean is 1/16*(1+2+...+16)=1/16*(136) = 17/2. Similarly, 1/16*(17+18+...+32)=1/16*(392)=49/2. So 49/2-17/2=32/2=16.
For the second problem, you have two equations and two unknowns. Let b be the number that you is the constant multiplied by the previous term. Then you have that
-2*b = x, which implies that b =-x/2
The second equation is that x*b = -32. Plugging b in from above and solving, you get x = positive or negative 8, and so b is positive or negative four. Then, using this, you can calculate that the sixth term is either
-2*4*4*4*4*4 =-2*4^5 = - 2048
or
-2*(-4)*(-4)*(-4)*(-4)*(-4) =-2*(-4)^5 = 2048
And so you see that 2048 is the correct answer.
16
imagine the numbers written in a line, with a line drawn along underneath them. The first mean is 1/4 of the way along, and the second mean is 3/4 of the way along. The distance between them is 1/2 of the length of the line. As the line has a total length of 32, the distance between them is 16
-2048
the series is -2 x 4^(n-1) where n is the term of the series
therefore -2 x 4^(6-1) = -2 x 1024 = -2048
16 for the first part
i think 512 for the second